'Football Crazy Hockey Mad' printed from http://nrich.maths.org/

We received a number of solutions for this problem with various
degrees of explanation and completeness. Some of them are below.
Well done everyone.

Adam from Bradfield School, Karina from The Manor Preparatory
School and Stefan from Ecclesfield School gave a solution to the
first part of the problem:

Each team wins two matches and loses two matches

Jack from Ecclesfield School offered some explanation for both
parts of the question, with a similar explanation offered by Matt
of Tapton School and Rebecca of The Chase High School:

The number of games possible has to be divisible by the number of
teams (editor`s comment: Why?). In the first league there are 10
possible games and 5 teams, so that is 2 games they have to win
each.

In the second league the number of games possible was 15 and the
number of teams was 6 so that wasn`t a whole number of games.

One solution for the first league would be:

Team 1 vs Team 2 (Team 1 won)

Team 1 vs Team 3 (team 1 won)

1 vs 4 (4 won)

1 vs 5 (5 won)

2 vs 3 (2 won)

2 vs 4 (2 won)

2 vs 5 (5 won)

3 vs 4 (3 won)

3 vs 5 (3 won)

4 vs 5 (4 won)

Matt from Tapton Comprehensive sent the folowing solution:

It is possible, for each of the five Football teams must win 2
games with same score and lose 2 games with the same score as that
with which they won.

The reason why you could have 5 teams is because there would be 10
matches and 5 teams and 10 divides by 5 and that equals 2 games
each that each team must win.

It would be impossible because each Hockey team could either win 0
games and lose 5 games, win 1 game and lose 4 games, win 2 games
and lose 3 games, win 3 games and lose 2 games, win 4 games and
lose 1 game or win all 5 games. But none of these would be even.
The reason why you could not have six teams leading because there
are 15 matches and 6 teams and 15 will not divide into 6
exactly.

Finally, a more general solution offered by Andrei of School 205
Bucharest:

I observed that, if there were 5 teams, there would be 10 matches
played. This result was obtained by adding all possible
combinations:

The first team has to play with all other 4 (4 matches)

The second team also plays 4 matches, but one was already taken
into account (3 matches).

The third team plays 4 matches, but two were already taken into
account (2 matches).

The fourth team plays 4 matches, but 3 were already taken into
account (1 match).

The fifth team plays 4 matches, but all were already taken into
account (no matches).

So, the total is: $4 + 3 + 2 + 1 = 4 \times5/2 = 10$
(matches).

In this case it is possible for all teams to be league leaders, as
it is possible for every team to win 2 matches, and to lose 2:

$10/5 = 2$ (won matches).

Now, if there are six teams, the total number of played matches
would be:
$$5 + 4 + 3 + 2 + 1 = 5 \times6/2 = 15$$
(matches).

As 15 is not divisible by 6, it`s impossible that every team will
win the same number of matches. So, it`s impossible for all the
teams to be league leaders.

Now, I look more generally at the problem:

If the number of teams is even ($2k$), than the total number of
matches would be: $$(2k - 1) + (2k - 2) + \dots + 1 = \frac{2k(2k -
1)}{2} = k(2k - 1)$$ This result should be divided by the number of
teams: $$\frac{k(2k - 1)}{2k} = \frac{(2k - 1)}{2}$$ This number
will never be divisible by 2, so, if there is an even number of
teams, not all of them could be league leaders.

Now, I analyze the case of an odd number of teams $(2k + 1)$; the
total number of matches is: $$2k + (2k - 1) + (2k - 2) + \dots + 1
= \frac{(2k + 1)2k}{2} = k(2k + 1)$$ This number should be divided
by the number of teams $(2k+1)$: $$\frac{k(2k + 1)}{2k + 1} = k$$
So, if the number of teams is odd, then all of them could be league
leaders.

Note: The total number of matches played in a round robin
tournament with $p$ teams, each playing only once with all others
is the number of ways of selecting a group of 2 teams from a group
of $p$, when order does not matter, i.e. combinations: $$^{p}C_{2}
= \frac{p!}{2!(p - 2)!} = \frac{p(p - 1)}{2}$$